Let $T\colon X\to X$ be an uppersemi Fredholm operator acting on a $B$space $X$ (the range of $T$ is closed and kernel is finitedimensional) with complemented range. Suppose $S\colon X\to X$ is bounded below. Does it follow that $T+S$ has complemented range? If we have assumed that $S$ is compact, then the answer would be 'yes'.

In fact, any bounded operator $L$ on $X$ may be written as a sum of two invertible operators $S$ and $T$, hence in particular both Fredholm and bounded below. We may take $S:=\lambda I$ and $T:=L\lambda I$, with $\lambda > \L\$. 


No. Take $U$ mapping $X$ isomorphically onto a subspace which is complemented via a projection $P$ and $V$ mapping $X$ isomorphically onto an uncomplemented subspace s.t. $PV=0$. (This situation is easily realizable in many spaces; e.g., having $X$ isomorphic to $X\oplus X$ and having $X$ contain an uncomplemented subspace isomorphic to itself is sufficient. Take for $U$ an isomorphism from $X\oplus X$ onto $X\oplus 0$ and let $V$ be an isomorphism from $X\oplus X$ onto an uncomplemented subspace of $0 \oplus X$ and let $P$ be the natural projection form $X\oplus X$ onto $X\oplus 0$.) Set $T=U+V$ and $S=U$. 

